Question

A long wire carrying a steady current is bent into a circle of single turn. The magnetic field at the centre of the coil is 'B'. If it is bent into a circular loop of radius 'r' having 'n' turns, the magnetic field at the centre of the coil for same current is ______.

विकल्प

• \[\frac {B}{n^2}\]

• \[\frac {B}{n}\]

• n²B

• nB

Answer 1

A long wire carrying a steady current is bent into a circle of single turn. The magnetic field at the centre of the coil is 'B'. If it is bent into a circular loop of radius 'r' having 'n' turns, the magnetic field at the centre of the coil for same current is n²B.

Explanation:

Magnetic field at the centre of single turn loop is B = \[\frac{\mu_{0}}{4\pi}\frac{2\pi\mathrm{I}}{\mathrm{r}}\]

\[\therefore\] Magnetic field at the centre of an n turn loop is

B_n = \[\left[{\frac{\mu_{0}}{4\pi}}{\frac{2\pi\mathrm{I}}{\left({\frac{r}{n}}\right)}}\right]\times\mathrm{n=n^{2}B}\]